傳遞函式引數（元組內的字典）

我正在撰寫一個函式來生成一個 API url，它接受多個引數并復制字串的一部分。簡短版本（只接受一個變數值Val）看起來像這樣

def get_url(par,att,value, _filter):

    base_url='''https://mytest.com/api/Test/''' str(par) '''?size=100000&filters=['''
    url = base_url   '''{"VarA" : "''' att '''", "Val":["''' value '''"], "VarB":"''' _filter '''"}]'''

    return url


print(get_url('Sales',{'Year':['2020'] , "Time"}))

這將回傳

https://mytest.com/api/Test/Sales?size=1000&filters=[{"VarA" : "Year", "Val":["2020"], "VarB":"Time"}]

我撰寫了另一個函式，以便我可以傳遞多個變數，Val同時還使用以下函式復制過濾器字串

def get_url(par,argv=None, _filter=None):
    if argv:
        att=argv
        url='''https://mytest.com/api/Test/''' str(par) '''?size=100000&filters=['''
        i=0
        for key in att.keys():
            attribute=key
            attribute_value=att[attribute]
            my_lst_str = '","'.join(map(str, attribute_value))
            values= '"' my_lst_str '"'
            if i==0:
                url=url '''{"VarA" : "''' attribute '''", "Val":[''' value '''], "VarB":''' str(_filter) '''}]'''
            else:url=url ''',{"VarA" : "''' attribute '''", "Val":[''' value '''], "VarB":''' str(_filter) '''}]'''
            i =1

    else:
        url = #Some string#

    return url


url = get_url('Sales',{'Loc':['USA','CAN'],'Year':["2016","2017"]},{'Country','Time'})

但是上面的代碼回傳的是這個

https://mytest.com/api/Test/Sales?size=1000&filters=[{"VarA" : "Loc", "Val":["USA","CAN"], "VarB":"{'Country', 'Time'}"}],{"VarA" : "Year", "Val":["2016","2017"], "VarB":"{'Country', 'Time'}"}]

我的預期輸出是

https://mytest.com/api/Test/Sales?size=1000&filters=[{"VarA" : "Loc", "Val":["USA","CAN"], "VarB":"Country"},{"VarA" : "Year", "Val":["2020","2017"], "VarB":"Time"}]

我如何隔離元素VarB？

uj5u.com熱心網友回復：

您可以使用您的索引：

from collections import OrderedDict
def get_url(par, argv=None, _filter=None):
    obj_list, obj_filter = [], list(_filter)
    if argv:
        att = argv
        url = (
            """https://mytest.com/api/Test/"""   str(par)   """?size=100000&filters="""
        )
        i = 0
        for key in att.keys():
            attribute = key
            attribute_value = att[attribute]
            obj_list.append({"VarA" : attribute, "Val": attribute_value, "VarB": obj_filter[i]})
            i  = 1

    else:
        url = "some url"

    return url str(obj_list)


url = get_url(
    "Sales", OrderedDict({"Loc": ["USA", "CAN"], "Year": ["2016", "2017"], "wewe": ["4", "5"]}), {"Country", "Time", "wre"}
)
print(url)

輸出：

https://mytest.com/api/Test/Sales?size=100000&filters=[{'VarA': 'Loc', 'Val': ['USA', 'CAN'], 'VarB': 'wre'}, {'VarA': 'Year', 'Val': ['2016', '2017'], 'VarB': 'Time'}, {'VarA': 'wewe', 'Val': ['4', '5'], 'VarB': 'Country'}]

uj5u.com熱心網友回復：

注意：集合中元素的順序是不確定的，盡管它可能由各種元素組成。通過將 a 轉換set為 a，list您無法確定訂單會保持不變！

>>> li = list({"Country", "Time", "w", "z"})
>>> li
['z', 'Country', 'w', 'Time']

問題中的代碼也無法運行。

我的回答基于您的預期輸出，我沒有考慮邊緣情況，我強烈建議更改函式引數。另外，我考慮_filter了一個串列，["Country", "Time"].

def get_url(par, att=None, _filter=None):
    url = f"https://mytest.com/api/Test/{par}?size=100000&filters="
    if att:
        result_list = [{'VarA':k, 'Val':v} for k , v in att.items()]
    else: return "some string"  

    if _filter:
        _filter.reverse()
        for d in result_list:
            d["VarB"] = _filter.pop()
    return f"{url}{result_list}"

url = get_url(
    "Sales", {"Loc": ["USA", "CAN"], "Year": ["2016", "2017"], "wewe": ["4", "5"]}, ["Country", "Time", "wre"]
)
print(url)

# https://mytest.com/api/Test/Sales?size=100000&filters=[{'VarA': 'Loc', 'Val': ['USA', 'CAN'], 'VarB': 'Country'}, {'VarA': 'Year', 'Val': ['2016', '2017'], 'VarB': 'Time'}, {'VarA': 'wewe', 'Val': ['4', '5'], 'VarB': 'wre'}]

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